What is the first law of thermodynamics?

Learn what the first law of thermodynamics is and how to use it.

What is the first law of thermodynamics?

Many power plants and engines operate by turning heat energy into work. The reason is that a heated gas can do work on mechanical turbines or pistons, causing them to move. The first law of thermodynamics applies the conservation of energy principle to systems where heat transfer and doing work are the methods of transferring energy into and out of the system. The first law of thermodynamics states that the change in internal energy of a system ΔU\Delta U equals the net heat transfer into the system QQ, plus the net work done on the system WW. In equation form, the first law of thermodynamics is,
ΔU=Q+W\Large \Delta U=Q+W
We have to be very careful with the first law. About half of textbooks, teachers, and professors write the first law of thermodynamics as ΔU=Q+W\Delta U=Q+W and the other half write it as ΔU=QW\Delta U=Q-W.
Both equations are correct, and they say the same thing. The reason for the difference is that in the formula ΔU=Q+Won gas\Delta U=Q+W_\text{on gas}, we are assuming that WW represents the work done on the system, and when we use ΔU=QWby gas\Delta U=Q-W_\text{by gas} we are assuming that WW represents the work done by the system.
The two different equations are equivalent since,
Won gas=Wby gasW_\text{on gas}=-W_\text{by gas}
When work is done on a system, the work done adds to the internal energy of the system (hence the plus sign in ΔU=Q+Won gas\Delta U=Q+W_\text{on gas}). When work is done by a system, the work done takes away from the internal energy of the system (hence the minus sign in ΔU=QWby gas\Delta U=Q-W_\text{by gas}).
We're going to use the equation ΔU=Q+Won gas\Delta U=Q+W_\text{on gas} with the plus sign on the work since that's the one used by the College Board AP physics 2 exam. This means that when we write WW, we will mean work done on the gas
Here ΔU\Delta U is the change in internal energy UU of the system. QQ is the net heat transferred into the system—that is, QQ is the sum of all heat transfer into and out of the system. WW is the net work done on the system.
So positive heat QQ adds energy to the system and positive work WW adds energy to the system. This is why the first law takes the form it does, ΔU=Q+W\Delta U=Q+W. It simply says that you can add to the internal energy by heating a system, or doing work on the system.

What do each of these terms (ΔU,Q,W\Delta U, Q, W) mean?

Nothing quite exemplifies the first law of thermodynamics as well as a gas (like air or helium) trapped in a container with a tightly fitting movable piston (as seen below). We'll assume the piston can move up and down, compressing the gas or allowing the gas to expand (but no gas is allowed to escape the container).
The gas molecules trapped in the container are the "system". Those gas molecules have kinetic energy.
Yes. Gas molecules can also have potential energy. Although we typically neglect the small change in gravitational potential energy due to the small changes in height of the gas in a container, a diatomic molecule like O2O_2 can have a potential energy associated with the oxygen atoms vibrating toward and away from each other, somewhat like two masses connected by a spring. This vibration "mode" of energy is typically not significant until temperatures get much higher than room temperature. At room temperature there simply isn't enough energy to get the atoms in a diatomic molecule to vibrate much.
Since most of the energy of a gas at reasonable temperatures will be in the form of kinetic energy, we can think of changes of internal energy as mostly being changes in the kinetic energy of the gas molecules (i.e. larger UU means faster moving gas molecules).
The internal energy UU of our system can be thought of as the sum of all the kinetic energies of the individual gas molecules. So, if the temperature TT of the gas increases, the gas molecules speed up and the internal energy UU of the gas increases (which means ΔU\Delta U is positive). Similarly, if the temperature TT of the gas decreases, the gas molecules slow down, and the internal energy UU of the gas decreases (which means ΔU\Delta U is negative).
It's really important to remember that internal energy UU and temperature TT will both increase when the speeds of the gas molecules increase, since they are really just two ways of measuring the same thing; how much energy is in a system. Since temperature and internal energy are proportional TUT \propto U, if the internal energy doubles the temperature doubles. Similarly, if the temperature does not change, the internal energy does not change.
One way we can increase the internal energy UU (and therefore the temperature) of the gas is by transferring heat QQ into the gas. We can do this by placing the container over a Bunsen burner or submerging it in boiling water. The high temperature environment would then conduct heat thermally through the walls of the container and into the gas, causing the gas molecules to move faster. If heat enters the gas, QQ will be a positive number. Conversely, we can decrease the internal energy of the gas by transferring heat out of the gas. We could do this by placing the container in an ice bath. If heat exits the gas, QQ will be a negative number. This sign convention for heat QQ is represented in the image below.
Since the piston can move, the piston can do work on the gas by moving downward and compressing the gas. The collision of the downward moving piston with the gas molecules causes the gas molecules to move faster, increasing the total internal energy. If the gas is compressed, the work done on the gas Won gasW_\text{on gas} is a positive number. Conversely, if the gas expands and pushes the piston upward, work is done by the gas. The collision of the gas molecules with the receding piston causes the gas molecules to slow down, decreasing the internal energy of the gas. If the gas expands, the work done on the gas Won gasW_\text{on gas} is a negative number. This sign convention for work WW is represented in the image below.
Below is a table that summarizes the signs conventions for all three quantities (ΔU,Q,W)(\Delta U, Q, W) discussed above.
ΔU\Delta U (change in internal energy)QQ (heat)WW (work done on gas)
is ++ if temperature TT increasesis ++ if heat enters gasis ++ if gas is compressed
is - if temperature TT decreasesis - if heat exits gasis - if gas expands
is 00 if temperature TT is constantis 00 if no heat exchangedis 00 if volume is constant

Is heat QQ the same thing as temperature T?T?

Absolutely not. This is one of the most common misconceptions when dealing with the first law of thermodynamics. The heat QQ represents the heat energy that enters a gas (e.g. thermal conduction through the walls of the container). The temperature TT on the other hand, is a number that's proportional to the total internal energy of the gas. So, QQ is the energy a gas gains through thermal conduction, but TT is proportional to the total amount of energy a gas has at a given moment. The heat that enters a gas might be zero (Q=0)(Q=0) if the container is thermally insulated, however, that does not mean that the temperature of the gas is zero (since the gas likely had some internal energy to start with).
To drive this point home, consider the fact that the temperature TT of a gas can increase even if heat QQ leaves the gas. This sounds counterintuitive, but since both work and heat can change the internal energy of a gas, they can both affect the temperature of a gas. For instance, if you place a piston in a sink of ice water, heat will conduct energy out the gas. However, if we compress the piston so that the work done on the gas is greater than the heat energy that leaves the gas, the total internal energy of the gas will increase.

What do solved examples involving the first law of thermodynamics look like?

Example 1: Nitrogen piston

A container has a sample of nitrogen gas and a tightly fitting movable piston that does not allow any of the gas to escape. During a thermodynamics process, 200 joules200\text{ joules} of heat enter the gas, and the gas does 300 joules300 \text{ joules} of work in the process.
What was the change in internal energy of the gas during the process described above?
Solution:
We'll start with the first law of thermodynamics.
ΔU=Q+W(start with the first law of thermodynamics)\Delta U=Q+W \quad \text{(start with the first law of thermodynamics)}
ΔU=(+200 J)+W(plug in Q=+200 J)\Delta U=(+200 \text{ J})+W \quad \text{(plug in }Q=+200\text{ J)}
Our convention is that the heat QQ will be a positive number if heat enters the gas, since it increases the internal energy of the gas.
ΔU=(+200 J)+(300 J)(plug in W=300 J)\Delta U=(+200 \text{ J})+(-300\text{ J}) \quad \text{(plug in }W=-300\text{ J)}
The convention we use is that the work is a positive number if work is done on the gas, since that adds energy to the gas. But since in this problem work was done by the gas, we plug in a negative number for the work done, since this subtracts energy from the gas.
ΔU=100 J(calculate and celebrate)\Delta U=-100\text{ J} \quad \text{(calculate and celebrate)}
Note: Since the internal energy of the gas decreases, the temperature must decrease as well.

Example 2: Heating helium

Four identical containers have equal amounts of helium gas that all start at the same initial temperature. Containers of gas also have a tightly fitting movable piston that does not allow any of the gas to escape. Each sample of gas is taken through a different process as described below:
Sample 1: 500 J500 \text{ J} of heat exits the gas and the gas does 300 J300 \text{ J} of work
Sample 2: 500 J500 \text{ J} of heat enters the gas and the gas does 300 J300 \text{ J} of work
Sample 3: 500 J500 \text{ J} of heat exits the gas and 300 J300 \text{ J} of work is done on the gas
Sample 4: 500 J500 \text{ J} of heat enters the gas and 300 J300 \text{ J} of work is done on the gas
Which of the following correctly ranks the final temperatures of the samples of gas after they're taken through the processes described above.
A. T4>T3>T2>T1T_4>T_3>T_2>T_1
B. T1>T3>T2>T4T_1>T_3>T_2>T_4
C. T4>T2>T3>T1T_4>T_2>T_3>T_1
D. T1>T4>T3>T2T_1>T_4>T_3>T_2
Solution:
Whichever gas has the largest increase in internal energy ΔU\Delta U will also have the greatest increase in temperature ΔT\Delta T (since temperature and internal energy are proportional). To determine how the internal energy changes, we'll use the first law of thermodynamics for each process.
Process 1:
ΔU=Q+WΔU=(500 J)+(300 J)ΔU=800 J\begin{aligned} \Delta U&=Q+W \\ \Delta U&=(-500\text{ J})+(-300\text{ J}) \\ \Delta U&=-800\text{ J} \end{aligned}
Process 2:
ΔU=Q+WΔU=(+500 J)+(300 J)ΔU=+200 J\begin{aligned} \Delta U&=Q+W \\ \Delta U&=(+500\text{ J})+(-300\text{ J}) \\ \Delta U&=+200\text{ J} \end{aligned}
Process 3:
ΔU=Q+WΔU=(500 J)+(300 J)ΔU=200 J\begin{aligned} \Delta U&=Q+W \\ \Delta U&=(-500\text{ J})+(300\text{ J}) \\ \Delta U&=-200\text{ J} \end{aligned}
Process 4:
ΔU=Q+WΔU=(+500 J)+(+300 J)ΔU=+800 J\begin{aligned} \Delta U&=Q+W \\ \Delta U&=(+500\text{ J})+(+300\text{ J}) \\ \Delta U&=+800\text{ J} \end{aligned}
The final temperatures of the gas will have the same ranking as the changes in internal energy (i.e. sample 4 has the largest increase in internal energy, so sample 4 will end with the largest temperature).
ΔU4>ΔU2>ΔU3>ΔU1\Delta U_4>\Delta U_2>\Delta U_3>\Delta U_1 and T4>T2>T3>T1T_4>T_2>T_3>T_1
So the correct answer is C.
This article was partially adapted from the following article:
  1. "The First Law of Thermodynamics" from Openstax College Physics. Download the original article free at http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@9.4:109/The-First-Law-of-Thermodynamic
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